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Zanzabum
3 years ago
9

A campground consists of 5 square campsites arranged in a line along a beach. The distance from the edge of a campsite to the wa

ter at the end of the beach is 4 yd. The area of the campground, including the beach, is 950 sq yd. What is the width of one campsite?
A. 14.35 yd
B. 13.93 yd
C. 11.93 yd

Mathematics
1 answer:
Anestetic [448]3 years ago
7 0
To answer this question with the choices given, you would substitute the possible answers into your work for the base and the height of each dimension on the entire campground.

To find the length of the campground, you multiply each choice by 5 because there are 5 campsites.

For the width, you add each choice to 4.

Then you multiply the total length by the total width to see if your answer is 950 square yards.

See the attached picture for all the work.

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In circle O, what is m∠MAJ? °
mr_godi [17]

See the attached figure.

m ∠KAJ = 170° &  m ∠LAM = 80°

We should know that :

m ∠KAJ + m ∠LAM + m ∠KAL + m ∠MAJ = 360°

∴ m ∠KAL + m ∠MAJ = 360° - (m ∠KAJ + m ∠LAM)

∴ m ∠KAL + m ∠MAJ = 360° - (170°+80°) = 360° - 250° = 110°

But : m ∠KAL = m ∠MAJ ⇒⇒⇒ <u>Opposite angles.</u>

∴ m ∠MAJ + m ∠MAJ = 110°

∴ 2 * m ∠MAJ = 110°

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Fantom [35]

Answer:

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C

Step-by-step explanation:

To find:

∫6x^5(x^6-2)\,dx

Solution:

Method of substitution:

Let x^6-2=t

Differentiate both sides with respect to t

6x^5\,dx=dt

[use (x^n)'=nx^{n-1}]

So,

∫6x^5(x^6-2)\,dx = ∫ t\,dt = \frac{t^2}{2}+C_1 where C_1 is a variable.

(Use ∫t^n\,dt=\frac{t^{n+1} }{n+1} )

Put t=x^6-2

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1

Use (a-b)^2=a^2+b^2-2ab

So,

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C

where C=2+C_1

Without using substitution:

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So, same answer is obtained in both the cases.

7 0
3 years ago
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